Page 226 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 19: The Open Mapping Theorem

Notes
Combining (i) – (iii), it follows that origin is also an interior point of (T(S )) . Consequently,
1
there exists > 0 such that
S T(S )
1
This justifies our claims.
We conclude the proof of the lemma by showing that
S /3 T(S ) , i.e., S T(S )
3
1
Let y B such that y < . Then y T(S ) and therefore there exists a vector x B such that
1 1
x < 1, y – y < /2 and y = T (x )
1 1 1 1
We next observe that

S /2 T(S 1/2 ) and y – y S /2
1
Therefore there exists a vector x B such that
2
1
x 3 , y y 1 y 2 2 and y = T (x )
2 2 2 2
Continuing this process, we obtain a sequence (x ) in B such that
n
1
x , y T(x ) and y (y y  y )
n n 1 n n 1 2 n 2
2 2
Let s = x + x + … + x , then
n 1 2 n
s = x + x + … + x
n 1 2 n
x + x + … + x
1 2 n
1 1 1
< 1 
2 2 2 2 n 1
1
2 1
2 n
< 2

Also for n > m, we have
s – s = s + … + x
n m m+1 n
x + … + x
m+1 n
1 1
< 
2 m 2 n 1
1 1
1
2 m 2 n m
= 1 (summing the G.P.)
1
2
1 1
= m 1 n m 1
2 2
0 as m, n

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